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Let $A$ be an $n\times n$ positive definite matrix. Show that there exists a unique positive definite matrix $B$ such that $B^2=A$.

I do know the existence. But what about the uniqueness? Would you help me out? Thank you.

Julien
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XLDD
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  • Can you diagonalize $B$? – Zach L. Apr 03 '13 at 03:14
  • See also http://math.stackexchange.com/questions/313564/square-root-of-a-real-matrix – Marc van Leeuwen Apr 04 '13 at 04:56
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    Check out the following for a clean proof - https://www.tandfonline.com/doi/pdf/10.1080/00207390500285867?casa_token=UV9s53_xQOQAAAAA:x7z1Ure9WGtjdQx2sXIoLQjyALUXhQ6A7itq4lAoiW9gA8zoZ7ET0FE_O7QzMzaFQ_DXzWhFbXPD - Koeber, Martin, and Uwe Schäfer. "The unique square root of a positive semidefinite matrix." International Journal of Mathematical Education in Science and Technology 37.8 (2006): 990-99 – David Veitch Oct 25 '20 at 21:41
  • https://christangdt.home.blog/2020/08/02/linear-algebra-2ed-hoffman-kunze-8-5/ see Soln of Q10. – R_Squared Nov 17 '22 at 09:00
  • Adding the condition of A being "Symmetric" is also necessary. See https://en.m.wikipedia.org/wiki/Square_root_of_a_matrix. See the theorem under the section "positive semidefinite matrices". – Avyaktha Achar Apr 27 '24 at 11:37

3 Answers3

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Being symmetric, $A$ is diagonalisable. Any square root $B$ commutes with $A$, so it must stabilise the eigenspaces of $A$. The restriction of $B$ to the eigenspace $V$ of $A$ for $\lambda>0$ is a symmetric positive definite square root of the restriction $\lambda I_V$ of $A$, so it suffices to show that $\sqrt\lambda I_V$ is the unique such square root.

But the restriction of $B$ is diagonalisable, any eigenvalue it has must be a square root of$~\lambda$ and also positive; a diagonalisable matrix with $\sqrt\lambda$ as unique eigenvalue cannot be other than $\sqrt\lambda I_V$.

Marc van Leeuwen
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Suppose $B_1=UDU^\ast$ and $B_2=V\Lambda V^\ast$ are two positive definite square roots of $A$, where $U$ and $V$ are unitary and $D,\Lambda$ are positive diagonal matrices. Since both $D$ and $\Lambda$ contain the positive square roots of the eigenvalues of $A$, the two matrices must be permutation similar. Therefore, by absorbing some appropriate permutation matrices into $U$ and $V$, we may assume WLOG that $D=\Lambda=(\lambda_1 I_{k_1})\oplus\cdots\oplus(\lambda_r I_{k_r})$, where $\lambda_1,\ldots,\lambda_r$ are distinct. Now the equality $B_1^2=B_2^2$ implies that $D$ commutes with $W=V^\ast U$. Hence $W$ must be a block diagonal matrix whose partitioning conforms to the block structure of $D$. But then $WDW^\ast=D$ and hence $B_1=UDU^\ast=VWDW^\ast V^\ast=VDV^\ast=B_2$.

user1551
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  • In my understanding, the key step here is that $W$ commutes with $D^2 \iff W$ commutes with $D$, am I right? – Avyaktha Achar Apr 27 '24 at 11:25
  • Proof :- $D=(\lambda_1 I_{k_1})\oplus\cdots\oplus(\lambda_r I_{k_r})$, hence $D^2=(\lambda_1^2 I_{k_1})\oplus\cdots\oplus(\lambda_r^2 I_{k_r})$, where $\lambda_1,\ldots,\lambda_r$ are distinct non-negative reals, and hence $\lambda_1^2,\ldots,\lambda_r^2$ are also distinct non-negative reals. As both $D$ and $D^2$ have the same block diagonal structure (with each block being a distinct multiple of some identity matrix), hence W commutes with $D$ iff its partitioning conforms to the block structure of $D$ iff its partitioning conforms to the block structure of $D^2$ iff it commutes with $D^2$. – Avyaktha Achar Apr 27 '24 at 11:27
  • Just to confirm, the proof that you gave works even if we replace "positive definite " by "positive semi definite" everywhere, am I right? – Avyaktha Achar Apr 27 '24 at 11:34
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    @AvyakthaAchar Yes. – user1551 Apr 27 '24 at 12:48
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Hints:

Can you write the matrix in Jordan Normal Form (why)?

If you have $B = P D^{1/2}P^{-1}$, where $D$ (of course, this can be complex valued) represents the square root of the diagonal matrix (which is just the square root of the eigenvalues), what does $B^2$ equal?

Amzoti
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